ParthKohli
  • ParthKohli
Let's try to correct this question.
Mathematics
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
Looks good to me! :-)
ParthKohli
  • ParthKohli
Here is the original question: Compute the sum.\[\int_{-4}^5 e^{(x+5)^2 }dx + 3\int_{1/3}^{2/3}e^{9(x-2/3)^2}dx \]
ParthKohli
  • ParthKohli
Pretty sure that the limits are the only thing wrong with the question (the result is beyond the scope of what is taught to us in high school). I'm trying to find out what the correct limits are. :)

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ParthKohli
  • ParthKohli
Now I see what is happening in the question. The exponent in the second integral is \((3x-2)^2\) and there is a 3 at the bottom. So we can substitute something like this:\[u +5= 3x-2\]\[u+5 = 2 - 3x\]
ParthKohli
  • ParthKohli
I'm going to post what I ended up with: had the limits on the second integral been 1/3 and -8/3, it would have been simpler.
ParthKohli
  • ParthKohli
Now I imagine that the guy who made the question tried to work backwards and didn't realise that the whole expression has to be positive. He first made the substitution\[u+5= 2-3x\]Which would land us with this expression:\[\int_{-4}^5 e^{(x+2)^2}dx - \int_{a}^{b}e^{(u+2)^2}du\]Now he wanted \(a\) to be \(-4\) and \(b\) to be \(5\).
ganeshie8
  • ganeshie8
How is that going to simplify anything ?
ParthKohli
  • ParthKohli
\[u = -3 - 3x\]For \(a\):\[a=-3-3(1/3)=-4\]Indeed, we get -4! For \(b\): \(5 = - 3 - 3x\) Basically, he's trying to find the original upper limit in the question (given as 2/3). Now he may have made a mistake like this: forgetting the whole negative sign.\[5 = 3 + 3x\]
ganeshie8
  • ganeshie8
you still need to use error function to evaluate the integral right ?
ParthKohli
  • ParthKohli
If you look carefully, \(a = -4\) and \(b = 5\) will make the whole expression zero.
ganeshie8
  • ganeshie8
Oh I see now...
ParthKohli
  • ParthKohli
Which is why 2/3 was mistakenly placed as the upper limit.
ganeshie8
  • ganeshie8
Nice! http://www.wolframalpha.com/input/?i=%5Cint_%7B-4%7D%5E5+e%5E%7B%28x%2B5%29%5E2+%7Ddx+%2B+3%5Cint_%7B1%2F3%7D%5E%7B-8%2F3%7De%5E%7B9%28x-2%2F3%29%5E2%7Ddx
ParthKohli
  • ParthKohli
Exactly. So the guy who framed the question made lots of mistakes.
ParthKohli
  • ParthKohli
Is there a way to get W|A to solve this equation?\[\int_{-4}^5 e^{(x+5)^2 }dx + 3\int_{1/3}^{t}e^{9(x-2/3)^2}dx=10\]I just arbitrarily chose 10 there. I want to solve for \(t\).
ParthKohli
  • ParthKohli
Hey I'm off to watching my favourite show :)
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=solve+1%2F2+sqrt%28pi%29+%28erfi%281%29-erfi%282-3+t%29%29%2B1%2F2+sqrt%28pi%29+%28erfi%2810%29-erfi%281%29%29+%3D+10
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=solve+%5Cint_%7B-4%7D%5E5+e%5E%7B%28x%2B5%29%5E2+%7Ddx+%2B+3%5Cint_%7B1%2F3%7D%5E%7Bt%7De%5E%7B9%28x-2%2F3%29%5E2%7Ddx%3D10
ganeshie8
  • ganeshie8
that looks really nasty...
ganeshie8
  • ganeshie8
what show ?

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